Computational Logic

Automated Deduction Fundamentals

Elements of First-Order Predicate Logic

First Order Language:

• An alphabet consists of the following classes of symbols:
  1. variables denoted by $X,Y,Z,Boo,...,$ (infinite)
  2. constants denoted by $1,a,boo,john,...,$
  3. functors denoted by $f,g,+,-,..,$
  4. predicate symbols denoted by $p,q,dog,...,$
  5. connectives, which are: $\neg$ (negation), $\vee$ (disjunction), $\wedge$ (conjunction), $\rightarrow$ (implication) and $\leftrightarrow$ (equivalence),
  6. quantifiers, which are: $\exists$ (there exists) and $\forall$ (for all),
  7. parentheses, which are: ( and ) and the comma, that is: ``,''.

• Each functor and predicate symbol has a fixed arity, they are often represented in $Functor/Arity$ form, e.g. f/3.

• A constant can be seen as a functor of arity 0.

• Propositions are represented by a predicate symbol of arity 0.

Important: Notation Convention Used

(A bit different from standard notational conventions in logic, but good for compatibility with LP systems)

 $$

Variables: start with a capital letter or a ``_'' (X, Y, _a, _1)

 $$

Atoms, functors, predicate symbols: start with a lower case letter or are enclosed in ' ' (f, g, a, 1, x, y, z, 'X', '_1')

Terms and Atoms

We define by induction two classes of strings of symbols over a given alphabet.

 $$

The class of terms:

• a variable is a term,
• a constant is a term,
• if $f$ is an $n$-ary functor and $t_1,...,t_n$ are terms then $f(t_1,..., t_n)$ is a term.

 $$

The class of atoms (different from LP!):

• a proposition is an atom,
• if $p$ is an $n$-ary pred. symbol and $t_1 ,...,t_n$ are terms then $p(t_1 ,...,t_n )$ is an atom,
• true and false are atoms.

 $$

The class of Well Formed Formulas (WFFs):

• an atom is a WFF,
• if $F$ and $G$ are WFFs then so are $\neg F, (F \vee G), (F \wedge G), (F \rightarrow G)$ and $(F \leftrightarrow G)$,
• if $F$ is a WFF and $X$ is a variable then $\exists X\ F$ and $\forall X\ F$ are WFF.

 $$

Literal: positive or negative (non-negated or negated) atom.

Examples

Examples of Terms

 $$

Given:

• constants: a, b, c, 1, spot, john...
• functors: f/1, g/3, h/2, +/3...
• variables: X, L, Y...

 $$

Correct: spot, f(john), f(X), +(1,2,3), +(X,Y,L), f(f(spot)), h(f(h(1,2)),L)

 $$

Incorrect: spot(X), +(1,2), g, f(f(h))

Examples of Literals

 $$

Given the elements above and:

• predicate symbols: dog/1, p/2, q/0, r/0, barks/1...

 $$

Correct: q, r, dog(spot), p(X,f(john))...

 $$

Incorrect: q(X), barks(f), dog(barks(X))

Examples (Contd.)

Examples of WFFs

 $$

Given the elements above

 $$

Correct: q, q $\rightarrow$ r, r $\leftarrow$ q, dog(X) $\leftarrow$ barks(X), dog(X) , p(X,Y), $\exists$ X (dog(X) $\wedge$ barks(X) $\wedge$ $\neg$ q), $\exists$ Y (dog(Y) $\rightarrow$ bark(Y))

 $$

Incorrect: q $\vee$, $\exists$ p

More about WFFs

• Allow us to represent knowledge and reason about it
  • Marcus was a man man(marcus)

  • Marcus was a pompeian pompeian(marcus)

  • All pompeians were romans $\forall$ X pompeian(X) $\rightarrow$ roman(X)

  • Caesar was a ruler ruler(caesar)

  • All romans were loyal to Caesar or they hated him
    $\forall$X roman(X) $\rightarrow$ loyalto(X,caesar) $\vee$ hate(X,caesar)

  • Everyone is loyal to someone $\forall$ X $\exists$ Y loyalto(X,Y)

• We can now reason about this knowledge using standard deductive mechanisms.

• But there is in principle no guarantee that we will prove a given theorem.

Towards Efficient Automated Deduction

• Automated deduction is search.

• Complexity of search: directly dependent on branching factor at nodes (exponentially!).

• It is vital to cut down the branching factor:
  • Canonical representation of nodes (allows identifying identical nodes).

  • As few inference rules as possible.

Towards Efficient Automated Deduction (Contd.)

Clausal Form

• The complete set of logical operators ( $\leftarrow, \wedge, \vee, \neg$,...) is redundant.

• A minimal (canonical) form would be interesting.

• It would be interesting to separate the quantifiers from the rest of the formula so that they did not need to be considered.

• It would also be nice if the formula were flat (i.e. no parenthesis).

• Conjunctive normal form has these properties [Davis 1960].

Deduction Mechanism

• A good example:
  Resolution - only two inference rules ( Resolution rule and Replacement rule).

Classical Clausal Form: Conjunctive Normal Form

• General formulas are converted to:
  
  • Set of Clauses.
  • Clauses are in a logical conjunction.
  • A clause is a disjunction of the form. $literal_1 \vee literal_2 \vee \dots \vee literal_n$
  • The $literal_i$ are negated or non-negated atoms.
  • All variables are implicitly universally quantified: i.e. if $X_1,...,X_k$ are the variables that appear in a clause it represents the formula:
    $\forall X_1,...,X_k\ \ \ literal_1 \vee literal_2 \vee \dots \vee literal_n$

• Any formula can be converted to clausal form automatically by:
  
  1. Converting to Prenex form.
  2. Converting to conjunctive normal form (conjunction of disjunctions).
  3. Converting to Skolem form (eliminating existential quantifiers).
  4. Eliminating universal quantifiers.
  5. Separating conjunctions into clauses.

• The unsatisfiability of a system is preserved.

Substitutions

• A substitution is a finite mapping from variables to terms, written as $\theta = \{ X_1/t_1,...,X_n/t_n \}$ where
  • the variables $X_1 ,...,X_n$ are different,
  • for $i=1,..., n \ X_i \neg\equiv t_i$.

• A pair $X_i/t_i$ is called a binding.

• $domain(\theta) = \{X_1,..,X_n\}$ and $range(\theta) = vars(\{t_1,...,t_n\})$.

• If $range(\theta) = \emptyset$ then $\theta$ is called ground.

• If $\theta$ is a bijective mapping from variables to variables then $\theta$ is called a renaming.

• Examples:
  • $\theta_1 = \{ X/f(A), Y/X, Z/h(b,Y), W/a\}$
  • $\theta_2 = \{ X/a, Y/a, Z/h(b,c), W/f(d)\}$ (ground)
  • $\theta_3 = \{ X/A, Y/B, Z/C, W/D\}$ (renaming)

Substitutions (Contd.)

• Substitutions operate on expressions, i.e. a term, a sequence of literals or a clause, denoted by $E$.

• The application of $\theta$ to $E$ (denoted $E \theta$) is obtained by simultaneously replacing each occurrence in $E$ of $X_i$ by $t_i$, $X_i/t_i \in \theta$.

• The resulting expression $E \theta$ is called an instance of $E$.

• If $\theta$ is a renaming then $E \theta$ is called a variant of $E$.

• Example:
  $\theta_1 = \{ X/f(A), Y/X, Z/h(b,Y), W/a\}$
  $p(X,Y,X) \ \theta_1 = p(f(A),X,f(A))$

Composition of Substitutions

• Given $\theta = \{ X_1 / t_1 ,...,X_n/t_n \}$ and $\eta = \{ Y_1/s_1 ,...,Y_m/s_m \}$ their composition $\theta \eta$ is defined by removing from the set
  
  $$\{ X_1/t_1 \eta ,...,X_n/t_n \eta , Y_1/s_1 ,...,Y_m/s_m \}$$
  
  
  those pairs $X_i/t_i \eta$ for which $X_i \equiv t_i \eta$, as well as those pairs $Y_i/s_i$ for which $Y_i \in \{ X_1,...,X_n \}$.

• Example: if $\theta = \{ X/3, Y/f(X,1) \}$ and $\eta = \{ X/4 \}$ then $\theta \eta = \{ X/3, Y/f(4,1) \}$.

• For all substitutions $\theta , \eta$ and $\gamma$ and an expression $E$

  i)

  $(E \theta ) \eta \equiv E( \theta \eta )$

  ii)

  $( \theta \eta ) \gamma = \theta ( \eta \gamma )$.

• $\theta$ is more general than $\eta$ if for some $\gamma$ we have $\eta = \theta \gamma$.

• Example: $\theta= \{X/f(Y)\}$ more general than $\eta= \{X/f(h(G))\}$

Unifiers

• If $A \theta \equiv B \theta$, then
  • $\theta$ is called a unifier of $A$ and $B$
  • $A$ and $B$ are unifiable

• A unifier $\theta$ of $A$ and $B$ is called a most general unifier (mgu) if it is more general than any other unifier of $A$ and $B$.

• If two atoms are unifiable then they have a most general unifier.

• $\theta$ is idempotent if $\theta \theta = \theta$.

• A unifier $\theta$ of $A$ and $B$ is relevant if all variables appearing either in $domain(\theta)$ or in $range(\theta)$, also appear in $A$ or $B$.

• If two atoms are unifiable then they have an mgu which is idempotent and relevant.

• An mgu is unique up to renaming.

Unification Algorithm

• Non-deterministically choose from the set of equations an equation of a form below and perform the associated action.
  
  1. $f(s_1 ,..., s_n ) = f(t_1 ,...,t_n )\rightarrow$ replace by $s_1 = t_1 ,...,s_n = t_n$
  2. $f(s_1 ,..., s_n ) = g (t_1 ,...,t_m )$ where $f \not\equiv g \rightarrow$ halt with failure
  3. $X = X \rightarrow$ delete the equation
  4. $t = X$ where t is not a variable $\rightarrow$ replace by the equation $X = t$
  5. $X = t$ where $X \not\equiv t$ and X has another occurrence in the set of equations $\rightarrow$

     5.1

     if X appears in t then halt with failure

     5.2

     otherwise apply $\{ X/t \}$ to every other equation

• Consider the set of equations $\{ f(x) = f(f(z)), g(a,y) = g(a,x) \}$:
  
  • (1) produces $\{ x = f(z), g(a,y) = g(a,x) \}$

  • then (1) yields $\{ x = f(z), a = a, y = x \}$

  • (3) produces $\{ x = f(z), y = x \}$

  • now only (5) can be applied, giving $\{ x = f(z), y = f(z) \}$

  • No step can be applied, the algorithm successfully terminates.

Unification Algorithm revisited

• Let $A$ and $B$ be two formulas:
  1. $\theta = \epsilon$
  2. while $A\theta \neq B\theta$:

     2.1

     find leftmost symbol in $A\theta$ s.t. the corresponding symbol in $B\theta$ is different

     2.2

     let $t_A$ and $t_B$ be the terms in $A\theta$ and $B\theta$ starting with those symbols

     (a)

     if neither $t_A$ nor $t_B$ are variables or one is a variable occurring in the other $\rightarrow$ halt with failure

     (b)

     otherwise, let $t_A$ be a variable $\rightarrow$ the new $\theta$ is the result of $\theta \{t_A/t_B\}$

  3. end with $\theta$ being an m.g.u. of $A$ and $B$

Unification Algorithm revisited (Contd.)

• Example: $A = p(X,X)$ $B = p(f(A),f(B))$

  
  $\theta$ $A\theta$ $B\theta$		Element 
  
  $\epsilon$		 $p(X,X)$ $p(f(A),f(B))$		$\{X/f(A)\}$ 
  
  $\{X/f(A)\}$ $p(f(A),f(A))$ $p(f(A),f(B))$		$\{A/B\}$
  
  $\{X/f(B),A/B\}$ $p(f(B),f(B))$ $p(f(B),f(B))$
  
  
  

• Example: $A = p(X,f(Y))$ $B = p(Z,X)$

  
  $\theta$ $A\theta$ $B\theta$		Element 
  
  $\epsilon$		 $p(X,f(Y))$ $p(Z,X)$		$\{X/Z\}$ 
  
  $\{X/Z\}$ $p(Z,f(Y))$ $p(Z,Z)$		$\{Z/f(Y)\}$
  
  $\{X/f(Y),Z/f(Y)\}$ $p(f(Y),f(Y))$ $p(f(Y),f(Y))$
  
  
  

Resolution with Variables

• It is a formal system with:
  • A first order language with the following formulas:
    • Clauses: without repetition, and without an order among their literals.
    • The empty clause $\Box$.

  • An empty set of axioms.
  • Two inference rules: resolution and replacement.

Resolution with Variables (Contd.)

• Resolution:

  $r_1$: $A \vee F_1 \vee \cdots \vee F_n$

  $r_2$: $\neg B \vee G_1 \vee \cdots \vee G_m$

  $((F_1 \vee \cdots \vee F_n)\sigma \vee G_1 \vee \cdots \vee G_m )\theta$

  where

  • $A$ and $B$ are unifiable with substitution $\theta$
  • $\sigma$ is a renaming s.t. $(A \vee F_1 \vee \cdots \vee F_n)\sigma$ and $\neg B \vee G_1 \vee \cdots \vee G_m$ have no variables in common
  • $\theta$ is the m.g.u. of $A\sigma$ and $B$
  The resulting clause is called the resolvent of $r_1$ and $r_2$.

• Replacement: $A \vee B \vee F_1 \vee \cdots \vee F_n \Rightarrow (A \vee F_1 \vee \cdots \vee F_n)\theta$ where
  • $A$ and $B$ are unifiable atoms
  • $\theta$ is the m.g.u. of $A$ and $B$

Basic Properties

• Resolution is correct - i.e. all conclusions obtained using it are valid.

• There is no guarantee of directly deriving a given theorem.

• However, resolution (under certain assumptions) is refutation complete:
  if we have a set of clauses $K = [C_0, C_1, \ldots, C_n]$ and it is inconsistent then resolution will arrive at the empty clause $\Box$ in a finite number of steps.

• Therefore, a valid theorem (or a question that has an answer) is guaranteed to be provable by refutation. To prove ``p'' given $K_0 = [C_0, C_1, \ldots, C_n]$:
  
  1. Negate it ($\neg p$).
  2. Construct $K = [\neg p, C_0, C_1, \ldots, C_n]$.
  3. Apply resolution steps repeatedly to K.

• Furthermore, we can obtain answers by composing the substitutions along a path that leads to $\Box$ (very important for realizing Greene's dream!).

• It is important to use a good method in applying the resolution steps - i.e. in building the resolution tree (or proof tree).

• Again, the main issue is to reduce the branching factor.

Proof Tree

• Given a set of clauses $K = \{C_0,C_1,\cdots,C_n\}$ the proof tree of $K$ is a tree s.t. :
  • the root is $C_0$
  • the branch from the root starts with the nodes labeled with $C_0,C_1,\cdots,C_n$

  • the descendent nodes of $C_n$ are labeled by clauses obtained from the parent clauses using resolution

  • a derivation in $K$ is a branch of the proof tree of $K$

• The derivation $C_0C_1 \cdots C_nF_0 \cdots F_m$ is denoted as $K,F_0\cdots F_m$

Proof Tree (Contd.)

• Example: part of the proof tree for K, with:
  

Characteristics of the Proof Tree

• It can be infinite:
  
  

• Even if it is finite, it can be too large to be explored efficiently

• Aim: determine some criteria to limit the number of derivations and the way in which the tree is explored $\Rightarrow$ strategy

• Any strategy based on this tree is correct: if $\Box$ appears in a subtree of the proof tree of $K$, then $\Box$ can be derived from $K$ and therefore $K$ is unsatisfiable

General Strategies

• Depth-first with backtracking: First descendant to the left; if failure or $\Box$ then backtrack
  

General Strategies (Contd.)

• Breadth first: all sons of all sibling nodes from left to right
  

General Strategies (Contd.) (Contd.)

• Iterative deepening
  • Advance depth-first for a time.
  • After a certain depth, switch to another branch as in breadth-first.

• Completeness issues / possible types of branches:
  • Success (always finite)
  • Finite failure
  • Infinite failure (provably infinite branches)
  • Non-provably infinite branches

Linear Strategies

• Those which only explore linear derivations
• A derivation $K,F_0\cdots F_m$ is linear if
  • $F_0$ is obtained by resolution or replacement using $C_0$
  • $F_i, i < 0$ is obtained by resolution or replacement using $F_{i-1}$

• Examples:
  
  

Characteristics of these Strategies

1

If $\Box$ can be derived from $K$ by using resolution with variables, it can also be derived by linear resolution

2

Let $K$ be $K\lq \cup \{C_0\}$ where $K\lq$ is a satisfiable set of clauses, i.e. $\Box$ cannot be derived from $K\lq$ by using resolution with variables. If $\Box$ can be derived from $K$ by using resolution with variables it can also be derived by linear resolution with root $C_0$.

 $$

From (1), if the strategy is breadth first, it is complete.

 $$

From (2), if we want to prove that $B$ is derived form $K\lq$ then we can apply linear resolution to $K = K\lq \cup \{\neg B\}$.

 $$

Depth first with backtracking is not complete:

Input Strategies

• Those which only explore input derivations
• A derivation $K,F_0\cdots F_m$ is input if
  • $F_0$ is obtained by resolution or replacement using $C_0$
  • $F_i, i < 0$ is obtained by resolution or replacement using at least a clause in $K$

• Example:
  
  

Input Strategies

• In an input derivation, if $F_{i-1}$ does not appear in any derivation of a successor clause, it can be eliminated from the derivation without changing the result

• If $F_{i-1}$ appears in the derivation of $F_j, j > 1$, $F_{i-1}$ can be allocated in position $j-1$

• As a result, we can limit ourselves to linear input derivations without losing any input derivable clause

• Let $K$ be $K\lq \cup \{C_0\}$ where $\Box$ is derived by using resolution with variables, $C_0$ is a negative Horn clause and all clauses in $K\lq$ are positive Horn clauses. There is an input derivation with root $C_0$ finishing in $\Box$ and in which the replacement rule is not used (Hernschen 1974)

• A Horn clause is a clause in which at most one literal is positive:
  • it is positive if precisely one literal is positive
  • it is negative if all literals are negatives

• As a result, in those conditions, a breadth first input strategy is complete, and a depth first input strategy with backtracking is complete if the tree is finite.

Ordered Strategies

• We consider a new formal system in which:
  
  1. clauses are ordered sets
  2. ordered resolution of two clauses
     $A = p_1 \vee \cdots \vee p_n\ and \ B = q_1 \vee \cdots \vee q_m$
     where $p_1$ is a positive literal and $q_1$ is a negative literal is possible iff $\neg p_1$ and $\sigma(q_1)$ are unifiable ($\sigma$ is a renaming, s.t. $p_1$ and $\sigma(q_1)$ have no variables in common)

  3. the resolvent of $A$ and $B$ is $\theta(p_2 \vee \cdots \vee p_n \vee \sigma(q_2 \vee \cdots \vee q_m))$ where $\theta$ is an m.g.u of $\neg p_1$ and $\sigma(q_1)$

• Let $K=K\lq \cup \{C_0\}$ be a set of clauses s.t. $\Box$ is derived by using resolution with variables, $C_0$ is a negative Horn clause and all clauses in $K\lq$ are positive Horn clauses with the positive literal in the first place. There is a sorted input derivation with root $C_0$ arriving at $\Box$.

• In this context a sorted linear input with:
  
  • breadth first: is complete
  • depth first with backtracking: is complete if the tree is finite